Avoid trailing zeroes in printf()
This can't be done with the normal printf
format specifiers. The closest you could get would be:
printf("%.6g", 359.013); // 359.013
printf("%.6g", 359.01); // 359.01
but the ".6" is the total numeric width so
printf("%.6g", 3.01357); // 3.01357
breaks it.
What you can do is to sprintf("%.20g")
the number to a string buffer then manipulate the string to only have N characters past the decimal point.
Assuming your number is in the variable num, the following function will remove all but the first N
decimals, then strip off the trailing zeros (and decimal point if they were all zeros).
char str[50];
sprintf (str,"%.20g",num); // Make the number.
morphNumericString (str, 3);
: :
void morphNumericString (char *s, int n) {
char *p;
int count;
p = strchr (s,'.'); // Find decimal point, if any.
if (p != NULL) {
count = n; // Adjust for more or less decimals.
while (count >= 0) { // Maximum decimals allowed.
count--;
if (*p == '\0') // If there's less than desired.
break;
p++; // Next character.
}
*p-- = '\0'; // Truncate string.
while (*p == '0') // Remove trailing zeros.
*p-- = '\0';
if (*p == '.') { // If all decimals were zeros, remove ".".
*p = '\0';
}
}
}
If you're not happy with the truncation aspect (which would turn 0.12399
into 0.123
rather than rounding it to 0.124
), you can actually use the rounding facilities already provided by printf
. You just need to analyse the number before-hand to dynamically create the widths, then use those to turn the number into a string:
#include <stdio.h>
void nDecimals (char *s, double d, int n) {
int sz; double d2;
// Allow for negative.
d2 = (d >= 0) ? d : -d;
sz = (d >= 0) ? 0 : 1;
// Add one for each whole digit (0.xx special case).
if (d2 < 1) sz++;
while (d2 >= 1) { d2 /= 10.0; sz++; }
// Adjust for decimal point and fractionals.
sz += 1 + n;
// Create format string then use it.
sprintf (s, "%*.*f", sz, n, d);
}
int main (void) {
char str[50];
double num[] = { 40, 359.01335, -359.00999,
359.01, 3.01357, 0.111111111, 1.1223344 };
for (int i = 0; i < sizeof(num)/sizeof(*num); i++) {
nDecimals (str, num[i], 3);
printf ("%30.20f -> %s\n", num[i], str);
}
return 0;
}
The whole point of nDecimals()
in this case is to correctly work out the field widths, then format the number using a format string based on that. The test harness main()
shows this in action:
40.00000000000000000000 -> 40.000
359.01335000000000263753 -> 359.013
-359.00999000000001615263 -> -359.010
359.00999999999999090505 -> 359.010
3.01357000000000008200 -> 3.014
0.11111111099999999852 -> 0.111
1.12233439999999995429 -> 1.122
Once you have the correctly rounded value, you can once again pass that to morphNumericString()
to remove trailing zeros by simply changing:
nDecimals (str, num[i], 3);
into:
nDecimals (str, num[i], 3);
morphNumericString (str, 3);
(or calling morphNumericString
at the end of nDecimals
but, in that case, I'd probably just combine the two into one function), and you end up with:
40.00000000000000000000 -> 40
359.01335000000000263753 -> 359.013
-359.00999000000001615263 -> -359.01
359.00999999999999090505 -> 359.01
3.01357000000000008200 -> 3.014
0.11111111099999999852 -> 0.111
1.12233439999999995429 -> 1.122
To get rid of the trailing zeros, you should use the "%g" format:
float num = 1.33;
printf("%g", num); //output: 1.33
After the question was clarified a bit, that suppressing zeros is not the only thing that was asked, but limiting the output to three decimal places was required as well. I think that can't be done with sprintf format strings alone. As Pax Diablo pointed out, string manipulation would be required.
I like the answer of R. slightly tweaked:
float f = 1234.56789;
printf("%d.%.0f", f, 1000*(f-(int)f));
'1000' determines the precision.
Power to the 0.5 rounding.
EDIT
Ok, this answer was edited a few times and I lost track what I was thinking a few years back (and originally it did not fill all the criteria). So here is a new version (that fills all criteria and handles negative numbers correctly):
double f = 1234.05678900;
char s[100];
int decimals = 10;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf("10 decimals: %d%s\n", (int)f, s+1);
And the test cases:
#import <stdio.h>
#import <stdlib.h>
#import <math.h>
int main(void){
double f = 1234.05678900;
char s[100];
int decimals;
decimals = 10;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf("10 decimals: %d%s\n", (int)f, s+1);
decimals = 3;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf(" 3 decimals: %d%s\n", (int)f, s+1);
f = -f;
decimals = 10;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf(" negative 10: %d%s\n", (int)f, s+1);
decimals = 3;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf(" negative 3: %d%s\n", (int)f, s+1);
decimals = 2;
f = 1.012;
sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals));
printf(" additional : %d%s\n", (int)f, s+1);
return 0;
}
And the output of the tests:
10 decimals: 1234.056789
3 decimals: 1234.057
negative 10: -1234.056789
negative 3: -1234.057
additional : 1.01
Now, all criteria are met:
- maximum number of decimals behind the zero is fixed
- trailing zeros are removed
- it does it mathematically right (right?)
- works (now) also when first decimal is zero
Unfortunately this answer is a two-liner as sprintf
does not return the string.